Problem 78
Question
Solve each equation. Identify each equation as a conditional equation, an inconsistent equation, or an identity. State the solution sets to the identities using interval notation. $$\frac{1}{x}+\frac{1}{x^{2}}=\frac{x+2}{x^{2}}$$
Step-by-Step Solution
Verified Answer
No solution; inconsistent equation.
1Step 1: Combine the fractions on the left side
Combine the two fractions on the left side by adding their numerators over the common denominator. equation: \(\frac{1}{x} + \frac{1}{x^{2}} = \frac{x+2}{x^{2}}\) becomes: \(\frac{x + 1}{x^{2}} = \frac{x + 2}{x^{2}}\)
2Step 2: Set the numerators equal
Since the denominators are equal, set the numerators equal to each other: \(x + 1 = x + 2\)
3Step 3: Solve for x
Subtract \(x\) from both sides of the equation to get: \(1 = 2\). This is a contradiction, so there is no solution.
4Step 4: Determine the type of equation
Since the equation leads to a contradiction, it is an inconsistent equation.
Key Concepts
conditional equationidentity equationinterval notation
conditional equation
A conditional equation is an equation that is true only under certain conditions or for specific values of the variables involved. Let's take an example: \(2x + 3 = 7\). To solve for \(x\), isolate the variable by performing operations that keep the equation balanced:
\[2x + 3 - 3 = 7 - 3 \] \[2x = 4 \] \[x = 2 \]
Here, \(x = 2\) is the specific condition that makes the equation true. If \(x\) takes any other value, the equation won't hold. Thus, a conditional equation depends on fulfilling certain criteria to be valid.
\[2x + 3 - 3 = 7 - 3 \] \[2x = 4 \] \[x = 2 \]
Here, \(x = 2\) is the specific condition that makes the equation true. If \(x\) takes any other value, the equation won't hold. Thus, a conditional equation depends on fulfilling certain criteria to be valid.
identity equation
An identity equation is true for all values of the variable(s) involved. This means that no matter what value you substitute for the variable, the equation will hold. For example, consider the equation: \[2(x + 3) = 2x + 6 \]
Let's distribute and simplify both sides:
\[2x + 6 = 2x + 6 \]
The simplified form shows that both sides are always equal, regardless of the value of \(x\). Identity equations provide a foundation for many algebraic principles and proofs, as they represent universal truths in mathematics.
Let's distribute and simplify both sides:
\[2x + 6 = 2x + 6 \]
The simplified form shows that both sides are always equal, regardless of the value of \(x\). Identity equations provide a foundation for many algebraic principles and proofs, as they represent universal truths in mathematics.
interval notation
Interval notation is a shorthand way of representing a set of numbers along a number line. It's used to describe solution sets, especially in inequalities. Let's break down the components:
\(1.\) **Brackets:** \[ \text{[ ]}\text{ - including the endpoint, e.g., [1, 5]} \] \[ \text{( )}\text{ - excluding the endpoint, e.g., (1, 5)} \]
\(2.\) **Intervals:** \[ \text{Finite Interval - Includes endpoints: [a, b] ; excludes endpoints: (a, b)} \] \[ \text{Infinite Interval - Open-ended: (a, ∞) or (-∞, b]} \]
Interval notation makes it easy to concisely express the range of values that solve an equation or inequality. For example, if \[ x > 3 \], we write this as \[ (3, \infty ) \]. This expresses that \(x\) can be any number greater than 3 but not 3 itself.
\(1.\) **Brackets:** \[ \text{[ ]}\text{ - including the endpoint, e.g., [1, 5]} \] \[ \text{( )}\text{ - excluding the endpoint, e.g., (1, 5)} \]
\(2.\) **Intervals:** \[ \text{Finite Interval - Includes endpoints: [a, b] ; excludes endpoints: (a, b)} \] \[ \text{Infinite Interval - Open-ended: (a, ∞) or (-∞, b]} \]
Interval notation makes it easy to concisely express the range of values that solve an equation or inequality. For example, if \[ x > 3 \], we write this as \[ (3, \infty ) \]. This expresses that \(x\) can be any number greater than 3 but not 3 itself.
Other exercises in this chapter
Problem 77
Perform the indicated operations. $$\frac{2 x^{2}+7 x-15}{4 x^{2}-100} \cdot \frac{2 x^{2}-9 x-5}{4 x^{2}-1}$$
View solution Problem 77
In place of each question mark in Exercises \(75-92,\) put an expression that will make the rational expressions equivalent. $$5=\frac{10}{?}$$
View solution Problem 78
Perform the indicated operations. $$\frac{x^{3}+1}{x^{2}-1} \cdot \frac{3 x-3}{x^{3}-x^{2}+x}$$
View solution Problem 78
In place of each question mark in Exercises \(75-92,\) put an expression that will make the rational expressions equivalent. $$\frac{3}{4}=\frac{12}{?}$$
View solution